Answer 3.3:4
From Förberedande kurs i matematik 2
(Difference between revisions)
m (Svar 3.3:4 moved to Answer 3.3:4: Robot: moved page) |
(Changed answer to 3.3:4c) |
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|c) | |c) | ||
| - | |width="50%"| <math>z= \left\{\begin{matrix} -1 \ | + | |width="50%"| <math>z= \left\{\begin{matrix} -1+i\sqrt{2} \\ -1-i\sqrt{2}\end{matrix}\right. </math> |
|d) | |d) | ||
|width="50%"| <math>z= \left\{\begin{matrix} (1+i\sqrt{15})/4\\ (1-i\sqrt{15})/4 \end{matrix}\right.</math> | |width="50%"| <math>z= \left\{\begin{matrix} (1+i\sqrt{15})/4\\ (1-i\sqrt{15})/4 \end{matrix}\right.</math> | ||
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Current revision
| a) | \displaystyle z= \left\{\begin{matrix} \phantom{-}(1+i)/\sqrt{2}\\ -(1+i)/\sqrt{2}\\ \end{matrix}\right. | b) | \displaystyle z = \left\{\begin{matrix} 2+i \\ 2-i \\ \end{matrix}\right. |
| c) | \displaystyle z= \left\{\begin{matrix} -1+i\sqrt{2} \\ -1-i\sqrt{2}\end{matrix}\right. | d) | \displaystyle z= \left\{\begin{matrix} (1+i\sqrt{15})/4\\ (1-i\sqrt{15})/4 \end{matrix}\right. |
